Can someone check where i have gone wrong in this.

$\displaystyle (\frac{u}{v})'= \frac{vu'-uv'}{v^2}$

$\displaystyle \frac{x^3y}{1-xy}$

$\displaystyle \frac{ [(1-xy) \cdot (3x^2y)] - [(x^3y) \cdot (-y)]} {(1-xy)^2} $

$\displaystyle = \frac{(3x^2y-3x^3y^2)-(-x^3y^2)}{(1-xy)^2} $

$\displaystyle =\frac{3x^2y-3x^3y^2+x^3y^2}{(1-xy)^2} $

$\displaystyle =\frac{3x^2y-2x^3y^2}{(1-xy)^2} $